cotpi 38 - Factors of factorials

[cotpi]

What is the largest integer n such that n! can not be expressed
as the sum of n distinct factors of itself?

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Originally posted at: http://cotpi.com/p/38/
Correct solutions will be archived at the URL mentioned above.

Solution to 'Splitting square into squares':
http://cotpi.com/p/36/#36-1

[Ilan Mayer]

SPOILER

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Largest such n is 2.

3! = 6 = 3 + 2 + 1

for 4! = 24 create 4 factors by multiplying the fisrt 2 above factors
by 4, and have 3 and 1 as additional factors, yielding 24 = 12+8+3+1

This can be continued for any n by multiplying the first n-2 factors
of the set for n-1 by n and adding n-1 and 1 as the last 2 factors.



Please reply to ilan dot mayer at hotmail dot com

__/\__
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__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
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[John Jones]

In article <jh5ksi$23u$1...@dont-email.me>, puz...@cotpi.com says...

>
> What is the largest integer n such that n! can not be expressed
> as the sum of n distinct factors of itself?

There is none largest.

Proof by induction according to the following scheme
(mathematical details left to the imagination)

3! = 3 + 2 + 1
4! = 12 + 8 + 3 + 1 (multiplying by 4 but splitting 4*1 as 3 + 1)
5! = 60 + 40+ 15 + 4 + 1 (multiplying by 5 but splitting 5*1 as 4 + 1)
etc

This works because if w divides n! then (n+1)*w divides (n+1)!
also (n+1)*w >= (n+1) > n and > 1
and (n+1) = n + 1, and n divides (n+1)! since n divides n!

so at stage n+1 the new list of numbers are indeed divisors, are n+1 in
number, and are distinct.

HTH
JJ

[John Jones]

In article <MPG.29a0b22bc...@news.eternal-september.org>,
a1...@hotamil.com says...
aha - didnt see Ilan's post and misread the problem
anyway ignore this in favour of ilan mayer's post.
JJ